Question

Use the results displayed in Problem 61 to predict the (a) magnitude and (b) inclination of Earth’s magnetic field at the geomagnetic equator, the (c) magnitude and (d) inclination at geomagnetic latitude $\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$, and the (e) magnitude and (f) inclination at the north geomagnetic pole.

Short answer

(a) Magnitude of Earth’s magnetic field at the geomagnetic equator is $3.10\times {10}^{-5}\text{T}$.

(b) Inclination of Earth’s magnetic field at the geomagnetic equator is $0°$.

(c) Magnitude of Earth’s magnetic field at the geomagnetic latitude $60°$is $5.59\times {10}^{-5}\text{T}$.

(d) Inclination of Earth’s magnetic field at the geomagnetic latitude $60°$is $73.9°$.

(e) Magnitude of Earth’s magnetic field at the north geomagnetic pole is $6.20\times {10}^{-5}\text{T}$.

(f) Inclination of Earth’s magnetic field at the north geomagnetic pole is $90°$.

Step-by-step solution

Understanding the concepts of magnetic field:

A magnetic field is a vector field that describes the magnetic effect on moving electric charges, electric currents, and magnetic materials. A moving charge in a magnetic field exerts a force perpendicular to its own velocity and to the magnetic field.

Here, you have to use the formula for Earth’s magnetic field,

$B=\frac{{\mu }_0\mu }{4\pi r^3}\times \sqrt{1+3{\sin }^2{\lambda }_m}$ ..... (1)

And the inclination of Earth’s magnetic field is,

$\tan \varphi =2\tan {\lambda }_m$ ..... (2)

Listing the known quantities:

Earth’s magnetic dipole moment,$\mu =8.00\times {10}^{22}\text{A}\cdot {\text{m}}^{\text{2}}$

The radius of the Earth, $r=6.378\times {10}^6\text{m}$

The permeability of free space,${\mu }_0=4\pi \times {10}^{-7}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^2$}\right.$

(a) Calculations of the magnitude of Earth’s magnetic field at the equator:

On equator, the wavelength is,

${\lambda }_m=0$

So the magnetic field using equation (1) is,

$\begin{array}{rcl}B& =& \frac{{\mu }_0\mu }{4\pi r^3}\times \sqrt{1+3{\sin }^2{\lambda }_m}\\ & =& \frac{(4\pi \times {10}^{-7}{\displaystyle \raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^2$}\right.})\times (8.00\times {10}^{22}\text{A}\cdot {\text{m}}^{\text{2}})}{4\pi \times {(6.378\times {10}^6\text{m})}^3}\times \sqrt{1+3{\sin }^2\left(0\right)}\\ & =& 3.10\times {10}^{-5}\text{T}\end{array}$

Hence, the magnitude of Earth’s magnetic field at the geomagnetic equator is $\begin{array}{rcl}& & 3.10\times {10}^{-5}\text{T}\end{array}$.

(b) Calculations of the inclination of Earth’s magnetic field at the equator

Use equation (2) to define the inclined of Earth's magnetic field at the equator.

$\begin{array}{rcl}\tan {\varphi }_i& =& 2\times \tan {\lambda }_m\\ & =& 2\times \tan 0°\end{array}$

${\varphi }_i=0°$

Hence, the inclination of Earth’s magnetic field at the geomagnetic equator is $0°$.

(c) Calculations of the Magnitude of Earth’s magnetic field at 60°:

The wavelength is,

${\lambda }_m=60$

Define the magnitude of Earth's magnetic field as below,

$\begin{array}{rcl}B& =& \frac{{\mu }_0\mu }{4\pi r^3}\times \sqrt{1+3{\sin }^2{\lambda }_m}\\ & =& \frac{(4\pi \times {10}^{-7}{\displaystyle \raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^2$}\right.})\times (8.00\times {10}^{22}\text{A}\cdot {\text{m}}^{\text{2}})}{4\pi \times {(6.378\times {10}^6\text{m})}^3}\times \sqrt{1+3{\sin }^2(60°)}\\ & =& 5.59\times {10}^{-5}\text{T}\end{array}$

Hence, the magnitude of Earth’s magnetic field at the geomagnetic latitude is $\begin{array}{rcl}& & 5.59\times {10}^{-5}\text{T}\end{array}$.

(d) Calculations of the inclination of Earth’s magnetic field at 60°.

Define the inclination of Earth’s magnetic field at the geomagnetic latitude is,

$\begin{array}{rcl}\tan {\varphi }_i& =& 2\times \tan {\lambda }_m\\ & =& 2\times \tan 60°\end{array}$

${\varphi }_i=73.9°$

Hence, the inclination of Earth’s magnetic field at the geomagnetic latitude $60°$is $73.9°$.

(e) Calculations of the magnitude of Earth’s magnetic field at the north pole:

The wavelength,

${\lambda }_m=90$

The magnitude of the Earth's magnetic field at north pole is,

$\begin{array}{rcl}B& =& \frac{{\mu }_0\mu }{4\pi r^3}\times \sqrt{1+3{\sin }^2{\lambda }_m}\\ & =& \frac{(4\pi \times {10}^{-7}{\displaystyle \raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^2$}\right.})\times (8.00\times {10}^{22}\text{A}\cdot {\text{m}}^{\text{2}})}{4\pi \times {(6.378\times {10}^6\text{m})}^3}\times \sqrt{1+3{\sin }^2(90°)}\\ & =& 6.20\times {10}^{-5}\text{T}\end{array}$

Hence, the magnitude of Earth’s magnetic field at the north geomagnetic pole is $\begin{array}{rcl}& & 6.20\times {10}^{-5}\text{T}\end{array}$.

(f) Calculations of the inclination of Earth’s magnetic field at the north pole:

Again use equation (2) to determine the inclination as below.

$\begin{array}{rcl}\tan {\varphi }_i& =& 2\times \tan {\lambda }_m\\ & =& 2\times \tan 90°\end{array}$

$\begin{array}{rcl}{\varphi }_i& =& 90°\end{array}$

Hence, the inclination of Earth’s magnetic field at the north geomagnetic pole is$90°$.